Updated 19 Oct 2026

JEE Main 2026, 21 Jan Shift 1 question paper with solutions

JEE Main 2026, 21 Jan Shift 1 was held on 21 January 2026. The bank holds 75 of its questions (Physics 25, Chemistry 25, Mathematics 25). A sample with answer keys and worked solutions is below, or sit the complete shift paper in the official NTA CBT interface (180 minutes, 300 marks, +4 / −1). Real PYQ papers are never filtered by your covered syllabus.

Sit this paper in the exam screen All JEE Main 2026 shifts Marks vs percentile table

2026 Exam session year

300 Maximum marks (+4 / −1)

180 min Official duration

75 Questions in bank for this shift

Try questions from JEE Main 2026, 21 Jan Shift 1

Q1PhysicsJEE Main 2026, 21 Jan Shift 1Atomic collisions

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If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is

____

m . (Atomic number of gold

= 79

and

\frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9}

in SI units)

Answer key (A) & worked solution

Correct answer: Option A

Energy conservation

K_{i} + U_{i} = K_{f} + U_{f} 7.7 \times 1 0^{6} \times 1.6 \times 1 0^{- 19} + 0 = 0 + \frac{9 \times 1 0 ^{9} ( 1.6 \times 1 0 ^{- 19} ) ( 79 \times 1.6 \times 1 0 ^{- 19} )}{r} r = 2.95 \times 1 0^{- 14}

Q2PhysicsJEE Main 2026, 21 Jan Shift 1Photoelectric effect and Einstein's photoelectric equation

A light wave described by

E = [60 sin (3 \times 1 0^{15}) t + sin (12 \times 1 0^{15})

t] (in SI units) falls on a metal surface of work function 2.8 eV . The maximum kinetic energy of ejected photoelectron is (approximately)

____

eV.

(h = 6.6 \times 1 0^{- 34} J - s .

and

e = 1.6 \times 1 0^{- 19} C

)

A5.1

B3.8

C6.0

D7.8

Answer key (A) & worked solution

Correct answer: Option A

ω_{1} = 3 \times 1 0^{15} rad / sec

ω_{2} = 12 \times 1 0^{15} rad / sec ∵ v = \frac{ω}{2 π} E_{photon} E_{m a x} K_{m a x} K_{m a x} = h v = 6.6 \times 1 0^{- 34} \times 1.91 \times 1 0^{15} = 1.26 \times 1 0^{- 18} J = \frac{1.26 \times 1 0 ^{- 18}}{1.6 \times 1 0 ^{- 19}} ≃ 7.9 eV = E_{m a x} - ϕ_{0} = 7.9 - 2.8 = 5.1 eV

Q3ChemistryJEE Main 2026, 21 Jan Shift 1Amino Acids

Identify the correct statements. A. Arginine and Tryptophan are essential amino acids. B. Histidine does not contain heterocyclic ring in its structure. C. Proline is a six membered cyclic ring amino acid. D. Glycine does not have chiral centre. E. Cysteine has characteristic feature of side chain as

MeS - CH_{2} - CH_{2} -

. Choose the correct answer from the options given below:

AC and E Only

BB and E Only

CC and D Only

DA and D Only

Answer key (D) & worked solution

Correct answer: Option D

- Histidine does contain heterocyclic ring. - Proline is a five membered cyclic ring amino acid. - Cysteine has characteristic feature of side chain as

CH_{2} - SMe

Q4ChemistryJEE Main 2026, 21 Jan Shift 1Miscellaneous examples

Given below are two statements: Statement I: The number of species among

SF_{4}, NH_{4}^{+}, [NiCl_{4}], XeF_{4}, [PtCl_{4}]^{2 -}, SeF_{4}

and

[Ni (CN)_{4}]^{2 -}

, that have tetrahedral geometry is 3 . Statement II: In the set

[NO_{2}, BeH_{2}, BF_{3}, AlCl_{3}]

, all the molecules have incomplete octet around central atom. In the light of the above statements, choose the correct answer from the options given below:

AStatement I is true but Statement II is false

BBoth Statement I and Statement II are false

CStatement I is false but Statement II is true

DBoth Statement I and Statement II are true

Answer key (C) & worked solution

Correct answer: Option C

Statement-I

SF_{4}

(See-saw)

XeF_{4}

(square planar),

[PtCl_{4}]^{2 -}

(square planar),

[NiCl_{4}]^{2 -}

(Tetrahedral),

[Ni (CN)_{4}]^{2 -}

(square planar),

SeF_{4}

(See-saw)

NH_{4}^{+}

(Tetrahedral) Statement-II

NO_{2}

(seven electrons on N )

BeH_{2}

(four electrons on Be )

BF_{3}

(six electrons on B )

AlCl_{3}

(six electrons on Al )

Q5MathematicsJEE Main 2026, 21 Jan Shift 1Equation reducible to quadratic

The sum of all the roots of the equation

(x - 1)^{2} - 5∣ x - 1∣ + 6 = 0

, is:

Answer key (A) & worked solution

Correct answer: Option A

Let

∣ x - 1∣ = t

t^{2} - 5 t + 6 = 0

t = 2& t = 3

∣ x - 1∣ = 2&∣ x - 1∣ = 3

x - 1 \pm 2& x - 1 = \pm 3

x = 1 \pm 2& x = 1 \pm 3

∴

root = 3, - 1, 4, - 2

∴

Sum of root

= 3 + (- 1) + 4 + (- 2) = 4

Q6MathematicsJEE Main 2026, 21 Jan Shift 1Cube roots of unity

x^{2} + x + 1 = 0

, then the value of

(x + \frac{1}{x})^{4} + (x^{2} + \frac{1}{x ^{2}})^{4} + (x^{3} + \frac{1}{x ^{3}})^{4} + \dots

+ (x^{25} + \frac{1}{x ^{25}})^{4}

is :

A128

B162

C175

D145

Answer key (D) & worked solution

Correct answer: Option D

x^{2} + x + 1 = 0

\Rightarrow x = ω

ω^{2}

∴ α = ω & β = ω^{2}

= (ω + ω^{2})^{4} + (ω^{2} + ω^{4})^{4} + (ω^{3} + ω^{6})^{4} + \dots + (ω^{25} + ω^{50})^{4}

= [(ω + ω^{2})^{4} + (ω^{2} + ω^{4})^{4} + (ω^{4} + ω^{8})^{4} + \dots . + (ω^{25} + ω^{50})^{4}] + [(ω^{3} + ω^{6})^{4} + (ω^{6} + ω^{12})^{4} + (ω^{9} + ω^{18})^{4} + \dots + (ω^{24} + ω^{48})^{4}]

= 17 times [1 + 1 + 1 \dots\dots + 1] + 8 times [(1 + 1)^{4} + (1 + 1)^{4} + \dots .. (1 + 1)^{4}]

= 17 + 128

= 145